c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)

\(P=\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\) với \(\dfrac{1}{4}< x< \dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}P=\sqrt{4x+2\sqrt{4x-1}}+\sqrt{4x-2\sqrt{4x-1}}\)

\(=\sqrt{\left(\sqrt{4x-1}\right)^2+2\sqrt{4x-1}+1}+\sqrt{\left(\sqrt{4x-1}\right)^2-2\sqrt{4x-1}+1}\)

\(=\sqrt{4x-1}+1+\left|\sqrt{4x-1}-1\right|\)

Do \(\dfrac{1}{4}< x< \dfrac{1}{2}\Leftrightarrow0< \sqrt{4x-1}< 1\)

\(\Rightarrow P=\dfrac{1}{\sqrt{2}}\left(\sqrt{4x-1}+1+1-\sqrt{4x-1}\right)=\sqrt{2}\)

Vậy \(P=\sqrt{2}\).

Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .

1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)

\(=>A=-12x+16\)

2) \(=>B=8x^3+27-8x^3+2=29\)

3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)

4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)

5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)

\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)

\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)

6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)

k cho mik nha , 

bạn đăng tách ra để mn cùng giúp nhé 

Bài 1 : 

a. (d) // (d') <=> \(\left\{{}\begin{matrix}m^2-2=-1\\3\ne m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=1\\m\ne1\end{matrix}\right.\Leftrightarrow m=-1\)

b, Hoành độ giao điểm (d1) ; (d2) tm pt 

\(2x+1=x+2\Leftrightarrow x=1\)

=> y = 3 

Vậy (d1) cắt (d2) tại A(1;3) 

Để 3 đường đồng quy khi (d3) đi qua A(1;3) 

hay A(1;3) thuộc (d3) 

<=> \(m^2+2-2m+1=3\Leftrightarrow m^2-2m=0\Leftrightarrow m=0;m=2\)

ta có :

\(8x\left(3x-8\right)+6x\left(-4x+7\right)=-88\)

\(\Leftrightarrow24x^2-64x-24x^2+42x=-88\Leftrightarrow-22x=-88\Leftrightarrow x=4\)

`8x (3x-8) +6x (-4x+7)=-88`

`-> 24x^2 - 64x - 24x^2 + 42x=-88`

`-> (24x^2 - 24x^2)+(-64x+42x)=-88`

`-> -22x=-88`

`->x=-88 : (-22)`

`->x=4`

Vậy `x=4`

5/8

3/4

\(a)\dfrac{3}{4}.\dfrac{5}{6}=\dfrac{5}{8}\)

\(b)\dfrac{2}{7}:\dfrac{4}{9}=\dfrac{9}{14}\)

\(\dfrac{3}{4}x\dfrac{5}{6}=\dfrac{3x5}{4x6}=\dfrac{15}{24}=\dfrac{5}{8}\)

b) \(\dfrac{2}{7}:\dfrac{4}{9}=\dfrac{2}{7}x\dfrac{9}{4}=\dfrac{18}{28}=\dfrac{9}{14}\)

\(y'=-3mx^2+2x-3\)

Hàm nghịch biến trên khoảng đã cho khi với mọi \(x\in\left(-3;0\right)\) ta có:

\(-3mx^2+2x-3\le0\)

\(\Leftrightarrow2x-3\le3mx^2\)

\(\Leftrightarrow\dfrac{2x-3}{3x^2}\le m\)

\(\Rightarrow m\ge\max\limits_{\left(-3;0\right)}\left(\dfrac{2x-3}{3x^2}\right)\)

Xét hàm \(f\left(x\right)=\dfrac{2x-3}{3x^2}\Rightarrow f'\left(x\right)=\dfrac{2\left(3-x\right)}{3x^3}< 0;\forall x\in\left(-3;0\right)\)

\(\Rightarrow f\left(x\right)>f\left(-3\right)=-\dfrac{1}{3}\)

\(\Rightarrow m\ge-\dfrac{1}{3}\)

CHọn B

Đặt \(x=\sqrt[3]{\sqrt[]{50}+7}-\sqrt[3]{\sqrt[]{50}-7}\)

\(x^3=14-3\sqrt[3]{\left(\sqrt[]{50}+7\right)\left(\sqrt[]{50}-7\right)}\left(\sqrt[3]{\sqrt[]{50}+7}-\sqrt[3]{\sqrt[]{50}-7}\right)\)

\(x^3=14-3x\)

\(x^3+3x-14=0\)

\(\left(x-2\right)\left(x^2+2x+7\right)=0\)

\(x=2\)

\(\Rightarrow\dfrac{m}{n}=2\)

\(\Rightarrow\) Hiển nhiên tồn tại vô số m, n nguyên thỏa mãn đẳng thức trên